(-1.5, 1.5)
(-1.5, 1.5)
Ingeniería Biomédica
2025-04-09
Let \(x[n]\) be a discrete-time signal.
The Z-Transform is defined as:
\[X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}\]
Where: - \(z \in \mathbb{C}\) is a complex variable - \(z = re^{j\omega}\)
Causal Signals
ROC is outside outermost pole
Anti-Causal Signals
ROC is inside innermost pole
\[H(z) = 1.00 \cdot \frac{(z - 0.50)}{(z - 0.90)}\]
(-1.5, 1.5)
(-1.5, 1.5)
If the ROC includes the unit circle, \(|z| = 1\), then:
\[X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n}\]
So the Fourier Transform is a special case of the Z-Transform.
Let:
\[x[n] = a^n u[n]\]
Where: - \(a \in \mathbb{R}\) - \(u[n]\) is the unit step function (0 for \(n<0\), 1 for \(n\geq 0\))
Apply the definition:
\[X(z) = \sum_{n=0}^{\infty} a^n z^{-n} = \sum_{n=0}^{\infty} (az^{-1})^n\]
This is a geometric series:
\[X(z) = \frac{1}{1 - az^{-1}} = \frac{z}{z - a}\]
A geometric series is a sum of terms where each term is multiplied by the same constant:
\[S = \sum_{n=0}^{\infty} r^n = 1 + r + r^2 + r^3 + \cdots\]
The value of \(r\) determines whether this sum converges (has a finite limit) or diverges.
Understand why this series:
\[\sum_{n=0}^{\infty} r^n\]
converges if and only if:
\[\boxed{|r| < 1}\]
Let’s consider the sum up to the \(N\)-th term:
\[S_N = \sum_{n=0}^{N} r^n = 1 + r + r^2 + \cdots + r^N\]
This has a closed-form expression:
\[S_N = \frac{1 - r^{N+1}}{1 - r} \quad \text{for } r \neq 1\]
To find the sum of the infinite series, take the limit as \(N \to \infty\):
\[S = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \frac{1 - r^{N+1}}{1 - r}\]
If \(|r| < 1\), then:
\[r^{N+1} \to 0 \quad \text{as } N \to \infty\]
So the sum becomes:
\[S = \frac{1}{1 - r}\]
✅ The geometric series converges.
❌ In all cases: the series diverges
\[S = 1 + 0.5 + 0.25 + 0.125 + \cdots = \frac{1}{1 - 0.5} = 2\]
Every term adds less. The sum “flattens out.”
The Z-Transform often gives us geometric series like:
\[\sum_{n=0}^{\infty} (az^{-1})^n\]
This converges only if:
\[|az^{-1}| < 1 \Rightarrow |z| > |a|\]
So, understanding convergence of geometric series = understanding ROC in Z-transforms.
| Condition | Behavior | Result |
|---|---|---|
| \(|r| < 1\) | Terms shrink | Series converges |
| \(|r| \geq 1\) | Terms grow or oscillate | Diverges |
\[\sum_{n=0}^{\infty} r^n = \frac{1}{1 - r} \quad \text{if } |r| < 1\]
For convergence of the geometric series:
\[|az^{-1}| < 1 \Rightarrow |z| > |a|\]
Therefore, the ROC is:
\[\boxed{|z| > |a|}\]
\[x[n] = (0.5)^n u[n]\]
Z-Transform:
\[X(z) = \sum_{n=0}^{\infty} (0.5)^n z^{-n} = \frac{z}{z - 0.5}\]
Region of Convergence:
\[\boxed{|z| > 0.5}\]
| Signal | Z-Transform | ROC |
|---|---|---|
| \(x[n] = a^n u[n]\) | \(\frac{z}{z - a}\) | \(|z| > |a|\) |
| Example: \(a = 0.5\) | \(\frac{z}{z - 0.5}\) | \(|z| > 0.5\) |
Let \(x[n] = a^n u[n]\), where \(|a| < 1\)
\[X(z) = \sum_{n=0}^{\infty} a^n z^{-n} = \frac{1}{1 - az^{-1}}, \quad \text{ROC: } |z| > |a|\]
A difference equation relates input and output values at different time steps.
\[y[n] - a_1 y[n-1] - a_2 y[n-2] = b_0 x[n] + b_1 x[n-1]\]
Common in: - Digital filters (FIR, IIR) - Signal models in ECG, EEG analysis - Implementation in real-time biosignal systems
The Z-Transform turns time shifts into powers of \(z^{-1}\):
| Time Domain | Z-Domain |
|---|---|
| \(x[n]\) | \(X(z)\) |
| \(x[n-k]\) | \(z^{-k} X(z)\) |
| \(y[n-k]\) | \(z^{-k} Y(z)\) |
Given:
\[y[n] - a_1 y[n-1] - a_2 y[n-2] = b_0 x[n] + b_1 x[n-1]\]
Apply \(\mathcal{Z} \{ \cdot \}\):
\[Y(z) - a_1 z^{-1} Y(z) - a_2 z^{-2} Y(z) = b_0 X(z) + b_1 z^{-1} X(z)\]
Group:
\[Y(z)(1 - a_1 z^{-1} - a_2 z^{-2}) = X(z)(b_0 + b_1 z^{-1})\]
Divide both sides:
\[H(z) = \frac{Y(z)}{X(z)} = \frac{b_0 + b_1 z^{-1}}{1 - a_1 z^{-1} - a_2 z^{-2}}\]
Given:
\[y[n] - 0.9 y[n-1] = x[n] - 0.5 x[n-1]\]
Z-Transform:
\[Y(z)(1 - 0.9 z^{-1}) = X(z)(1 - 0.5 z^{-1})\]
Transfer Function:
\[H(z) = \frac{1 - 0.5 z^{-1}}{1 - 0.9 z^{-1}}\]
Let’s analyze \(H(z)\):
Pole-Zero Plot
Visualizes system behavior
Check for: - Stability (poles inside unit circle) - Frequency shaping
Convert this equation:
\[y[n] = 0.6 y[n-1] + x[n] + x[n-1]\]
Find: - \(H(z)\) - Poles and zeros - Plot them in the Z-plane